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Let $a$ and $d$ be positive integers. We assume $(-a)/d=-a/d$. By definition, $a=dq+r$, where $q$ and $r$ are arbitrary integers such that $0\leq r <d.$ Thus, $a/d=q+r/d$ with $0 \leq r/d < 1.$ We can substitute both $(-a)/d$ and $-a/d$ into floor functions:

$\lfloor(-a)/d\rfloor=\lfloor-q-r/d\rfloor=\begin{cases} -q &\text{if } r=0 \\ -q-1 &\text{if } r>0 \end{cases}$ and

$-\lfloor a/d \rfloor=-\lfloor q+r/d \rfloor = -q$.

Thus, we can say that $(-a)/d=-a/d$ when $r=0$. So $a/d=q$. Therefore, $d$ divides $a$.

Conversely, if $d$ divides $a$, then we can write $a$ in the form $dq$ for some arbitrary integer $q$. So, $(-a)/d=(-dq)/d=-q$ and $-a/d=-dq/d=-q$. Thus, $(-a)/d=-a/d.$

So we have proven that if $a$ and $d$ are positive integers, then $(-a)/d=-a/d$ if and only if $d$ divides $a.$

Is this proof correct? If so, are there any improvements I could make?

Gabriel Dylan
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  • Yes, it is correct. "We assume $(-a)/d=-a/d$" was a bit confusing as now you're using both real number division and integer division in the proof. Also, if this is the usual real number division, I don't see the need to assume. You could simply refer to it like "For real number division, we know that $(-a)/d=-a/d$"

    Also, maybe it's redundant, but you may want to explicitly link integer division to the floor function like, $$a ; \text{div} ; d \equiv \lfloor a/d \rfloor$$

     – Tony Mathew Sep 26 '23 at 04:24

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